Classical Mechanics Problem 1: Central Potential Solution...Classical Mechanics Problem 1: Central...

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Classical Mechanics Problem 1: Central Potential Solution a) Integrals of motion for a central potential V (r): Angular Momentum L = rv t = r 2 ˙ φ Energy per unit mass E = 1 2 ( ˙ r 2 + v 2 t ) + V (r)= 1 2 ˙ r 2 + V eff (r) where v t is the tangential velocity and V eff is defined as V eff (r)= V (r)+ L 2 2r 2 If the orbit is circular, the distance of the test body from the origin is invariant: ˙ r = 0, which implies that the body is always at the equilibrium-distance: dV eff dr =0 dV dr = L 2 r 3 = v 2 t r = r ˙ φ 2 then ˙ φ = ω φ = L r 2 = 1 r dV dr 1/2 so for the period we get T φ = 2π ω φ =2π 1 r dV dr -1/2 b) Write the orbit as in the statement of the problem: r(t)= r 0 + ²(t) with dV eff dr (r 0 )=0 and ² 2 ¿ r 2 0 . The energy per unit mass is now E = 1 2 ˙ ² 2 + V eff (r 0 + ²), and since ² is small we may Taylor-expand the potential as V eff (r 0 + ²)= V eff (r 0 )+ dV eff dr (r 0 )² | {z } =0 + 1 2 d 2 V eff dr 2 (r 0 )² 2 + O(² 3 ) so then E - V eff (r 0 )= 1 2 ˙ ² 2 + 1 2 d 2 V eff dr 2 (r 0 )² 2 + O(² 3 ) = const. In the above equation we readily recognize the equation of the simple harmonic oscil- lator with ω r = d 2 V eff dr 2 1/2 r=r0 1

Transcript of Classical Mechanics Problem 1: Central Potential Solution...Classical Mechanics Problem 1: Central...

  • Classical Mechanics Problem 1: Central PotentialSolution

    a) Integrals of motion for a central potential V (r):

    Angular Momentum L = rvt = r2φ̇Energy per unit mass E = 12

    (ṙ2 + v2t

    )+ V (r) = 12 ṙ

    2 + Veff(r)

    where vt is the tangential velocity and Veff is defined as

    Veff(r) = V (r) +L2

    2r2

    If the orbit is circular, the distance of the test body from the origin is invariant: ṙ = 0,which implies that the body is always at the equilibrium-distance:

    dVeffdr

    = 0 ⇒ dVdr

    =L2

    r3=

    v2tr

    = rφ̇2

    then

    φ̇ = ωφ =L

    r2=

    (1r

    dV

    dr

    )1/2

    so for the period we get

    Tφ =2πωφ

    = 2π(

    1r

    dV

    dr

    )−1/2

    b) Write the orbit as in the statement of the problem:

    r(t) = r0 + ²(t) withdVeffdr

    (r0) = 0 and ²2 ¿ r20.

    The energy per unit mass is now E = 12 ²̇2 + Veff(r0 + ²), and since ² is small we may

    Taylor-expand the potential as

    Veff(r0 + ²) = Veff(r0) +dVeffdr

    (r0)²︸ ︷︷ ︸

    =0

    +12

    d2Veffdr2

    (r0)²2 +O(²3)

    so then

    E − Veff(r0) = 12 ²̇2 +

    12

    d2Veffdr2

    (r0)²2 +O(²3) = const.In the above equation we readily recognize the equation of the simple harmonic oscil-lator with

    ωr =(

    d2Veffdr2

    )1/2

    r=r0

    1

  • and its general solution is

    ²(t) =

    √E − Veff(r0)

    ωrcos [ωr(t− t0)]

    where t0 is an arbitrary constant.

    Now return to writing ωr in terms of V (r) instead of Veff(r).

    ω2r =d2Veffdr2

    =d2V

    dr2+

    3L2

    r4=

    d2V

    dr2+ 3ω2φ =

    d2V

    dr2+

    3r

    dV

    dr

    ωr =(

    d2V

    dr2+

    3r

    dV

    dr

    )1/2

    r=r0

    =[

    1r3

    d

    dr

    (r3

    dV

    dr

    )]1/2

    r=r0

    And the radial period is

    Tr =2πωr

    c) Stability is determined by the sign of ω2r . For stability: ω2r > 0, so

    1r3

    d

    dr

    (r3

    dV

    dr

    )> 0

    for the Yukawa-potential

    V (r) = −GMr

    e−kr

    so the condition is

    1r3

    d

    dr

    (r3

    dV

    dr

    )=

    GM

    r3e−kr

    [1 + kr − (kr)2] > 0 ⇒ [1 + kr − (kr)2] > 0

    [1 + kr − (kr)2] =

    (√5− 12

    + kr

    ) (√5 + 12

    − kr)

    > 0

    which is satisfied only if

    kr <

    (√5 + 12

    )

    Therefore circular orbits are unstable for

    kr >

    (√5 + 12

    )

    ¥

    2

  • d) The outermost stable circular orbit is at

    r0 =

    (√5 + 12k

    )

    its energy per unit mass is

    E = V (r0) +12(r0ωφ)2 = V (r0) +

    12

    (rdV

    dr

    )

    r=r0

    12

    GM

    r0e−kr0(kr0 − 1) = GM

    r0e−kr0

    (√5− 14

    )> 0

    If r0 is decreased only slightly, E > 0 still and the orbit is absolutely stable ¥The effective potential for the Yukawa-potential has the form shown in Figure 1.

    r

    Veff

    Figure 1: Effective potential against distance from the origin

    3

  • Classical Mechanics Problem 2: Planar Double PendulumSolution

    l

    l

    q1

    q2

    a) L = T − VThe moment of inertia for a uniform rod of length l and mass m is

    I =13ml2 about one of the ends

    andIc =

    112

    ml2 about the rod’s center

    The kinetic energy term we can decompose into three parts:

    T = T1 + T2,rot + T2,trans

    where T1 is the kinetic energy of the first rod, T2,trans is the translational energy ofthe center of mass of the second rod and T2,rot is its rotational energy about its centerof mass. Then

    T1 =16ml2θ̇21

    T2,rot =124

    ml2θ̇22

    andT2,trans =

    12m

    (ẋ2c + ẏ

    2c

    )

    where xc and yc are the coordinates of the second rod’s center of mass, so

    xc = l sin θ1 +l

    2sin θ2

    yc = −l cos θ1 − l2 cos θ2

    from which

    ẋ2c + ẏ2c = l

    2

    [θ̇21 +

    14θ̇22 + θ̇1θ̇2 (sin θ1 sin θ2 + cos θ1 cos θ2)

    ]

    4

  • The potential energies are simply Vi = mgyc,i, where yc,i are the vertical coordinates ofthe rods’ centers of mass. Since both rods are uniform, yc,i are simply the coordinatesof the centers. Thus,

    V1 = −mg l2 cos θ1; V2 = −mg(

    l cos θ1 +l

    2cos θ2

    )

    The full Lagrangian is then

    L = T1 + T2,rot + T2,trans − V1 − V2

    = ml2[23θ̇21 +

    16θ̇22 + θ̇1θ̇2 cos (θ1 − θ2)

    ]+ mgl

    [32

    cos θ1 +12

    cos θ2

    ]

    b) Expand the Langrangian from part a) for small angles. The only function we have todeal with is

    cos θ = 1− 12θ2 +O(θ4)

    Since we are going to look for normal modes with θj = θ̂j exp (iωt), where the θ̂j ¿ 1,we immediately see that in the term θ̇1θ̇2 cos (θ1 − θ2), the θ-dependence in the cosinecan be dropped, because even the first θ-dependent term gives a fourth order correction.Then the approximate Lagrangian is

    L = ml2[23θ̇21 +

    16θ̇22 + θ̇1θ̇2

    ]−mgl

    [34θ21 +

    14θ22

    ]+ const.

    The Euler-Lagrange equations are

    d

    dt

    ∂L

    ∂θ̇j=

    ∂L

    ∂θj

    so in the specific case:

    43θ̈1 +

    12θ̈2 +

    g

    l

    (32θ1

    )= 0

    12θ̈1 +

    13θ̈2 +

    g

    l

    (12θ2

    )= 0

    if we now look for normal modes, as mentioned, the above set of equations takes theform [ (

    43ω

    2 − 32 gl)

    12ω

    2

    12ω

    2(

    13ω

    2 − 12 gl)

    ] [θ̂1

    θ̂2

    ]= 0

    Non-trivial solutions exist if the determinant of the matrix on the left is zero. Denotingω2 = λg/l, we can write this condition as

    (43λ− 3

    2

    ) (13λ− 1

    2

    )− λ

    2

    4= 0,

    5

  • that is736

    λ2 − 76λ +

    34

    = 0

    whose solutions areλ± = 3± 6√

    7,

    so finally

    ω± =[(

    3± 6√7

    )g

    l

    ]1/2

    c) To sketch the eigenmodes, find eigenvectors of the matrix in part b).

    • ω2 = λ−g/l (low-frequency mode)

    θ̂2 =(

    3λ− 8

    3

    )θ̂1 =

    13

    (2√

    7− 1)

    θ̂1

    (2√

    7− 1)/3 > 0 and real, therefore the two pendula are in phase;• ω2 = λ+g/l (high-frequency mode)

    θ̂2 =(

    3λ− 8

    3

    )θ̂1 =

    13

    (−2√

    7− 1)

    θ̂1

    (−2√7− 1)/3 < 0 and real, therefore the two pendula are perfectly out of phase.

    a b

    Figure 2: The low- (a) and high-frequency (b) normal modes of the planar double pendulum.

    6

  • Electromagnetism Problem 1Solution

    a) Normal modes are products of harmonic standing waves in the x, y and z directions.For their frequencies, we have

    ω = c√

    k2x + k2y + k2z = c[(πnx

    a

    )2+

    (πnyb

    )2+

    (πnzb

    )2]1/2; nx, ny, nz ∈ Z+

    Since a > b, the lowest frequency has nx = 1 and either ny = 1, nz = 0 or ny = 0, nz = 1(note that ny = 0, nz = 0 does not satisfy the boundary condition E‖,at wall = 0). Sincewe are told to pick the mode with ~E ‖ ŷ, the boundary conditions require

    ~E(r, t) = E0ŷ sinπx

    asin

    πz

    bcosωt

    The magnetic induction we can get from Faraday’s Law:

    ∂ ~B

    ∂t= −c

    (∇× ~E

    )= x̂c

    ∂Ey∂z

    − ẑc∂Ey∂x

    =(x̂cE0

    π

    bsin

    πx

    acos

    πz

    b− ẑcE0 π

    acos

    πx

    asin

    πz

    b

    )cosωt

    ~B(r, t) = −πcE0ω

    (x̂

    bsin

    πx

    acos

    πz

    b− ẑ

    acos

    πx

    asin

    πz

    b

    )sin ωt

    where the frequency ω is (by the argument above)

    ω = πc(

    1a2

    +1b2

    )1/2

    b) At a boundary of media, the discontinuity in the normal component of the electric fieldis 4π times the surface charge density σ, so

    Ey(x, 0, z) = 4πσ

    σ(x, 0, z) =E04π

    sinπx

    asin

    πz

    bcosωt

    σ(x, b, z) = −σ(x, 0, z)and

    σ(0, y, z) = σ(a, y, z) = σ(x, y, 0) = σ(x, y, b) ≡ 0Similarly, at the boundary of media the discontinuity of the tangential component ofthe magnetic field is given by the surface current ~κ

    n̂× ~B = 4πc

    7

  • where n̂ is a unit vector normal to the surface, so

    ~κ(x, 0, z) =c2E04ω

    (ẑ

    bsin

    πx

    acos

    πz

    b+

    acos

    πx

    asin

    πz

    b

    )sin ωt

    ~κ(x, b, z) = −~κ(x, 0, z)

    ~κ(0, y, z) = −c2E04ω

    asin

    πz

    bsin ωt

    ~κ(a, y, z) = −~κ(0, y, z)

    ~κ(x, y, 0) = −c2E04ω

    bsin

    πx

    asinωt

    ~κ(x, y, b) = −~κ(x, y, 0)

    c) Since there is no charge on the b × b sides, the force there is purely magnetic and isgiven by

    ~F (t) =12c

    b×b

    (~κ× ~B

    )d2x

    ~F (x = 0, t) = −E20c

    8ω2a2x̂

    b∫

    0

    dy

    b∫

    0

    dz sin2πz

    bsin2ωt

    = −x̂(

    c

    4ωb

    aE0 sin ωt

    )2

    ~F (x = a, t) = −~F (x = 0, t)The forces point outwards from the box on both sides (as is indicated by the sign inthe equation above).

    d) Start with the sides where y = const. The magnetic component of the force can bewritten as above

    ~Fmag(y = 0, t) = −E20c

    8ω2ŷ

    a∫

    0

    dx

    b∫

    0

    dz

    (1a2

    cos2πx

    asin2

    πz

    b+

    1b2

    sin2πx

    acos2

    πz

    b

    )sin2ωt

    = −ŷ 12

    ( c2ω

    E0 sin ωt)2 1

    (b

    a+

    a

    b

    )

    To simplify this result further, use ω from part a)

    ω−2 = (πc)−2(

    1a2

    +1b2

    )−1= ab(πc)−2

    (b

    a+

    a

    b

    )−1

    Then~Fmag(y = 0, t) = −ŷ

    (E04

    sin ωt)2

    ab

    2π3

    8

  • The electric component of the force can be written as

    ~Fel(y = 0, t) =∫

    12σ ~E d2x = ŷ

    E208π

    cos2ωt

    a∫

    0

    dx

    b∫

    0

    dz sin2πx

    asin2

    πz

    b

    = ŷ12

    (E04

    cos ωt)2

    ab

    π3

    ~Ftot(y = 0, t) = ~Fel(y = 0, t) + ~Fmag(y = 0, t)

    = ŷ(

    E04

    )2ab

    2π3(cos2ωt− sin2ωt)

    = ŷ(

    E04

    )2ab

    2π3cos 2ωt

    and~Ftot(y = b, t) = −~Ftot(y = 0, t)

    There net force on the top and bottom sides oscillates between the inward and out-ward direction with half the period of the lowest frequency mode. In a time average,therefore, this force cancels.

    Next, calculate the force on the sides where z = const. Again, there is no charge,therefore no electric component; the force is purely magnetic

    ~F (z = 0, t) = −E20c

    8ω2b2ẑ

    b∫

    0

    dy

    a∫

    0

    dx sin2πx

    asin2ωt

    = −ẑ ab

    ( c4ω

    E0 sin ωt)2

    ~F (z = b, t) = −~F (z = 0, t)

    The magnetic force is pushing the a× b walls outwards, too (sign!).e) From the Maxwell stress tensor, the force per unit surface area is

    ~f =14π

    ~E( ~E · n̂)− E2

    8πn̂ +

    14π

    ~B( ~B · n̂)− B2

    8πn̂

    On the x = const. walls n̂ = ±x̂, ~E = 0 and ~B · x̂ = 0, so

    ~f(x = {0, a}, t) = ∓B2

    8πx̂ = ∓E

    20c

    8ω2a2x̂ sin2

    πz

    bsin2ωt

    which is exactly the integrand from part c).

    9

  • On the y = const. walls n̂ = ±ŷ, ~E( ~E · ŷ) = E2ŷ and ~B · ŷ = 0, so we get

    ~f(y = {0, b}, t) = ∓ 18π

    (E2 −B2)ŷ

    = ±[E208π

    cos2ωt sin2πx

    asin2

    πz

    b

    −E20c

    8ω2

    (1a2

    cos2πx

    asin2

    πz

    b+

    1b2

    sin2πx

    acos2

    πz

    b

    )sin2ωt

    ]ŷ

    the sum of the first two integrands from part d).

    On the z = const. walls n̂ = ±ẑ, ( ~E · ẑ) = 0 and ~B · ẑ = 0, so we get

    ~f(z = {0, b}, t) = ∓B2

    8πẑ = ∓E

    20c

    8ω2b2ẑ sin2

    πx

    asin2ωt

    the last integrand from part d).

    F(z)

    k(y)

    k(z)

    k(x)

    x

    b

    b

    a

    z

    y

    s

    –s

    k(y)k(z)

    k(x)

    F(x)

    F(x)F(z)

    Figure 3: Average total forces, surface charges and surface currents on the cavity.

    10

  • Electromagnetism Problem 2: Waves in a Dilute GasSolution

    (see Feynman Lectures on Physics, vol. II, chapter 32)

    a) The EM wave is travelling in the x̂ direction; it has a transverse electric field, so assumeE × ŷ = 0. Then the electron in the atom behaves classically as a damped, drivenharmonic oscillator

    me(ÿ + γẏ + ω20y

    )= −qE0e−iωt

    with the solution

    y(t) =1

    ω2 − ω20 + iγωqE(t)me

    .

    For the dipole moment per unit volume:

    P = na(−q)y = 1ω20 − ω2 − iγω

    naq2E

    me

    Therefore the volume polarizability is, according to the definition given,

    α(ω) =P

    ²0E=

    1ω20 − ω2 − iγω

    naq2

    ²0me

    (A quantum mechanical derivation would give this same expression multiplied by theoscillator strength f for the transition.)

    b) With no free charges or currents, Maxwell’s equations read

    ∇ ·D = 0; ∇×E = −∂B∂t

    ∇ ·B = 0; ∇×H = ∂D∂t

    and B = µ0H, D = ²0E + P = ²0(1 + α)E for a single frequency ω. This gives us thefollowing wave-equation

    ∂2D∂t2

    − 1µ0²0(1 + α)

    ∇2D = 0.

    Now let D be that of a plane wave: D ∝ ei(kx−ωt). Then

    k2 = µ0²0(1 + α)ω2 = (1 + α)ω2

    c2

    ⇒ n(ω) =√

    1 + α(ω)

    One can also get this result by using the microscopic E, B and P fields:

    ∇ ·E = − 1²0∇ ·P; c2∇×B = ∂

    ∂t

    (P²0

    + E)

    11

  • ⇒ ∂2E

    ∂t2− c2∇2E = − 1

    ²0

    ∂P∂t

    also∂2P∂t2

    + γ∂P∂t

    + ω20P = −naq

    2

    meE.

    Together these give us k2 = (1 + α)ω2/c2 for a plane wave, as before. (Note that weare neglecting dipole-dipole interactions in the dilute gas.)

    c) We start by noting that according to Fourier-analysis

    E(x, t) =12π

    ∞∫

    −∞dk ei(kx−ω(k)t)Ê(k)

    Ê(k) =

    ∞∫

    −∞dx e−ikxE(x, 0) =

    1√2πσ2

    ∞∫

    −∞dx e−i(k−kc)x−x

    2/(2σ2)

    = e−i(k−kc)2σ2/2

    Now Taylor-expand ω(k) about k = kc:

    ω(k) = ω(kc) +(

    dk

    )

    kc

    (k − kc) +O{(k − kc)2}

    ≡ kcvph + vg(k − kc) +O{(k − kc)2}where, by definition, vph and vg are the phase- and group-velocities, respectively. Nowlet K = k − kc. Then

    E(x, t) =12π

    ∞∫

    −∞dK eikc(x−vpht)+iK(x−vgt)−K

    2σ2/2

    E(x, t) = eikc(x−vpht)e(x−vgt)

    2/(2σ2)

    √2πσ2

    = eikc(x−vpht)N(x− vgt, σ)

    d) From part c)

    vg =dω

    dk=

    (dk

    )−1=

    c

    n

    (1 +

    d log nd log ω

    )−1.

    For the dilute gas, n =√

    1 + α ≈ 1 + α/2, which we will write as n = nr + ini (for αis complex)

    nr ≈ 1 + naq2

    2²0meω20 − ω2

    (ω20 − ω2)2 + γ2ω2

    ni ≈ naq2

    2²0meγω

    (ω20 − ω2)2 + γ2ω2

    12

  • Here the real part nr of the index of refraction determines the dispersion, and theimaginary part ni determines the absorption/gain coefficient. At ω = ω0:

    nr = 1 andd log nrd log ω

    = − naq2

    2²0meγ2

    vg =(

    1− naq2

    2²0meγ2

    )−1c ≈

    (1 +

    naq2

    2²0meγ2

    )c

    Note that vg > c at ω = ω0. This is called anomalous dispersion. It does not vio-late causality because signals (information) cannot travel faster than the minimum of(vph, vg), and now vph = c (since nr = 1). Also, the waves are damped by the electronicresonance maximally at ω = ω0.

    13

  • Quantum Mechanics Problem 1Solution

    1. The ground state will have no nodes, so we can pick the even part of the general solutionof the free Schrödinger equation inside the well. Outside the well, square-integrabilitydemands the solutions to vanish at infinity. The wave-function for the ground state isthen

    |x| < w |x| > wψ(x) = cos kx ψ(x) = Ae−α|x|

    Both the wave-function and its derivative has to be continuous at the boundaries ofthe well:

    ψ : cos kw = Ae−αw

    dx: −k sin kw = −αAe−αw

    ⇒ k tan kw = α

    Directly from Schrödinger’s equation:

    |x| < w |x| > w

    E =~2k2

    2mE =− ~

    2α2

    2m+ V0

    ⇒ ~2k2

    2m= −~

    2α2

    2m+ V0

    From which we get the transcendental equation:

    k tan kw =[2mV0~2

    − k2]1/2

    Let k∗ denote the positive root of the equation above, and introduce the followingnotation:

    kc =√

    2mV0~

    and kmax =π

    2w.

    Clearly, the LHS of the equation diverges at kmax; and the RHS describes a circle withradius kc, as shown in Fig. 4.

    For the energy we have

    E =~2k∗2

    2m

    14

  • *kc kmax

    k

    k*

    k*

    k** tan * w

    kc

    k tankw

    H kc2- k2 L 1/ 2( – )

    Figure 4: Graphical representation of the solution of the transcendental equation

    2. Write the result of part 1 in the non-dimensional form:

    kw tan kw =[2mw2V0~2

    − (kw)2]1/2

    According to the condition given in the statement of the problem, the radius of thecircle on the RHS (that in Fig. 4) goes to infinity, therefore

    k → kmaxand

    E → ~2k2max2m

    =~2π2

    8mw2

    3. The potential barrier on the low-potential side of the well, denoted a in the figure, willbe finite (for any E), so the particle will eventually escape by the tunnel-effect.

    4. ∆E = 0, because the perturbation is odd (and therefore its integral with the square ofthe ground-state wave-function vanishes).

    5.

    F =1~

    a∫

    w

    dx√

    2m(V (x)−B)

    V (x) = V0 − eEx

    B = V0 − eEa ⇒ a = V0 −BeE

    15

  • x

    V x( )

    V0

    –w w0

    a

    F =√

    2m~

    a∫

    w

    dx√

    (V0 −B)− eEx

    =√

    2m~

    23

    (− 1

    eE)

    (V0 −B − eEx)3/2∣∣∣∣a

    w

    =√

    2m~

    23

    1eE (V0 −B − eEw)

    3/2

    6. Write the energy of the particle as

    B =12mv2.

    Thenv2 =

    2Bm

    .

    The time it takes for the particle to bounce back and forth once is

    T =4wv

    ,

    so it hits the right wall with frequency

    ν =v

    4w

    ⇒ Probability to escapeunit time

    =v

    4we−2F

    ⇒ Lifetime ∼ 4wv

    e2F

    16

  • Quantum Mechanics Problem 2Solution

    1. Drop the t-label for simplicity. Then we have

    H = B[

    cos θ sin θe−iωt

    sin θeiωt − cos θ]

    ,

    and for the eigenvectors solve[

    cos θ sin θe−iωt

    sin θeiωt − cos θ] [

    x

    y

    ]= ±

    [x

    y

    ]

    with the normalization condition |x|2 + |y|2 = 1. From the vector-equation

    y = eiωtsin θ

    cos θ ± 1x

    and with the normalization, we end up with

    |+〉 =[

    cos (θ/2)sin (θ/2)eiωt

    ]|−〉 =

    [sin (θ/2)

    − cos (θ/2)eiωt]

    2. Decompose the state-vector as

    |ψ〉 = c+|+〉+ c−|−〉and write Schrödinger’s equation in terms of these vectors:

    i~d

    dt|ψ〉 = H|ψ〉

    i~[ċ+|+〉+ c+ d

    dt|+〉+ ċ−|−〉+ c− d

    dt|−〉

    ]= B [c+|+〉 − c−|−〉]

    or in the (|+〉, |−〉) basis

    i~d

    dt

    [c+

    c−

    ]=

    [B − i~〈+| ddt |+〉 −i~〈+| ddt |−〉−i~〈−| ddt |+〉 −B − i~〈−| ddt |−〉

    ][c+

    c−

    ]

    which, with the given concrete form of the vectors, is

    i~d

    dt

    [c+

    c−

    ]=

    [B + ~ω sin2(θ/2) −~ω cos (θ/2) sin (θ/2)

    −~ω sin (θ/2) cos (θ/2) −B + ~ω cos2(θ/2)

    ][c+

    c−

    ].

    Now use the identities

    sin2(θ/2) =12(1− cos θ)

    cos2(θ/2) =12(1 + cos θ)

    sin (θ/2) cos (θ/2) =12

    sin θ

    17

  • to get

    i~d

    dt

    [c+

    c−

    ]=

    12

    [2B − ~ω cos θ −~ω sin θ−~ω sin θ −2B + ~ω cos θ

    ][c+

    c−

    ].

    Note that in the last equation we dropped the part of the Hamiltonian that was pro-portional to the identity, since that gives only a time dependent phase that is identicalfor the coefficients c−, c+. This we can rewrite in the form:

    i~d

    dt

    [c+

    c−

    ]=

    [Dz Dx

    Dx −Dz

    ][c+

    c−

    ].

    with the solution[

    c+

    c−

    ]= exp

    {−i~

    [Dz Dx

    Dx −Dz

    ]}[1

    0

    ]

    = cos

    (| ~D|t~

    )− iD̂ · ~σ sin

    (| ~D|t~

    ).

    And so

    c+ = cos

    (| ~D|t~

    )− i Dz

    | ~D|sin

    (| ~D|t~

    )

    |c+|2 = cos2(| ~D|t~

    )− iD

    2z

    D2sin2

    (| ~D|t~

    )

    3. For B À ~ω, Dz → D, so|c+|2 → 1

    (Adiabatic theorem)

    18

  • Statistical Mechanics and Thermodynamics Problem 1Thermodynamics of a Non-Interacting Bose Gas

    Solution

    a)

    np =1

    eβ(Ep−µ) − 1 Ep =p2

    2m

    At and below TBEC µ = 0. At exactly TBEC, there are no atoms in the condensate and

    N =V

    2π~3

    ∫d3p

    eβp2/(2m) − 1 = (2π~)−3

    V

    (2mβ

    )3/24π

    ∞∫

    0

    x2dx

    ex2 − 1︸ ︷︷ ︸

    =I1

    n =1

    2π2

    (2mkT~2

    )3/2I1

    kTBEC =(

    2π2nI1

    )2/3 ~22m

    (2 points)

    b) The above integral with µ = 0 also applies below TBEC, but it then gives the numberof non-condensed atoms. So on an isotherm below Vcritical

    • Nnon-condensed is constant• T is constant⇒ p is constant

    (think of the kinetic origin of pressure)

    V

    p

    Classical Gas 1/p Vµ

    Phase Transition Line

    pBEC

    µ V–5/3

    Bose Gas

    (2 points)

    19

  • c)

    U =V

    2π~3

    ∫p2

    2md3p

    eβp2/(2m) − 1 = (2π~)−3

    V

    (2mβ

    )5/2 4π2m

    ∞∫

    0

    x4dx

    ex2 − 1︸ ︷︷ ︸

    =I2

    = NcI2I1

    (2mβ

    )1

    2m= Nc

    I2I1

    kT ∝ T 5/2

    cv =52

    I2I1

    Nck =V

    2π2

    (2mkT~2

    )3/2k

    (52I2

    )∝ T 3/2

    (2 points)

    d) From the reversibility of the Carnot-cycle:

    dS1 = −dS2 for 1 cycle∆S1 = −∆S2 for the entire process

    dU = TdS − pdV︸︷︷︸=0

    ⇒ T ∂S∂T

    ∣∣∣∣V

    =∂U

    ∂T

    ∣∣∣∣V

    = cv = aT 3/2︸ ︷︷ ︸from c)

    ⇒ dSdT

    =cvT

    Therefore the entropy transfer in the entire process is

    ∆Si =

    T0∫

    Ti

    cvT

    dT = a

    T0∫

    Ti

    T 1/2dT =23a

    (T

    3/20 − T 3/2i

    )

    ∆S1 + ∆S2 = 0 ⇒ T 3/20 =12

    (T

    3/21 + T

    3/22

    )

    Heat transferred to F1 : Q1 =

    T0∫

    T1

    TdS =25a

    (T

    5/20 − T 5/21

    ).

    Heat transferred from F2 : Q2 =

    T2∫

    T0

    TdS =25a

    (T

    5/22 − T 5/20

    ).

    Therefore the total work done by the Carnot-machine is

    W = Q2 −Q1 = 25a(T

    5/21 + T

    5/22 − 2T 5/20

    )

    (4 points)

    20

  • Statistical Mechanics and Thermodynamics Problem 2Phase Transition in a Superconductor

    Solution

    a)

    cH ≡ ∂Q∂T

    ∣∣∣∣H

    = T∂S

    ∂T

    ∣∣∣∣H

    dS =∂S

    ∂T

    ∣∣∣∣M

    dT +∂S

    ∂M

    ∣∣∣∣T

    dM

    ∂S

    ∂T

    ∣∣∣∣H

    =∂S

    ∂T

    ∣∣∣∣M

    +∂S

    ∂M

    ∣∣∣∣T

    ∂M

    ∂T

    ∣∣∣∣H︸ ︷︷ ︸

    =0

    where the last term is zero because M is independent of T . Then

    cH = T∂S

    ∂T

    ∣∣∣∣M

    =∂Q

    ∂T

    ∣∣∣∣M

    ≡ cM

    (2 points)

    b) The transition takes place at constant T and H. The thermodynamic function whosevariables are T and H is the Gibbs-potential:

    dG = −SdT −MdHGsuper = Gnormal at every point on HC(T ), so dGS = dGN which we then write as

    −SSdT −MSdH = −SNdT − MN︸︷︷︸=0

    dH

    dH

    dT

    ∣∣∣∣trans.line

    =dHCdT

    =SN − SS

    MS= − 4π

    V HC(T )(SN − SS)

    (3 points)

    c) By the third law S → 0 as T → 0. But the figure shows HC(T = 0) is finite. Therefore

    dHCdT

    → 0 as T → 0.

    The transition is second order where SN − SS = 0, that is, the latent heat equals zero.

    SN − SS = − V4πHC(T )dHCdT

    21

  • • At T = 0 the transition is second order because both entropies go to zero.• At T = TC(H = 0) the transition is second order since HC(T ) = 0 and dHC/dT

    is finite.

    • At all other temperatures the transition is first order since both HC(T ) anddHC/dT are finite.

    (2 points)

    d) Use H and T as variables

    dS(H,T ) =∂S

    ∂H

    ∣∣∣∣T

    dH +∂S

    ∂T

    ∣∣∣∣H

    dT

    ∂S

    ∂H

    ∣∣∣∣T

    =cHT

    ∂S

    ∂T

    ∣∣∣∣H

    =↑

    Maxwellrelation

    − ∂M∂T

    ∣∣∣∣H

    = 0

    S =∫

    cHT

    dT =a

    3T 3V T < TC

    =b

    3T 3V + γTV T > TC

    SN − SS =(

    b− a3

    )T 3V + γTV T = TC(H = 0)

    γ =(

    b− a3

    )T 2C

    TC(H = 0) =(

    3γb− a

    )1/2

    (3 points)

    22